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A053289
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First differences of consecutive perfect powers (A001597).
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10
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3, 4, 1, 7, 9, 2, 5, 4, 13, 15, 17, 19, 21, 4, 3, 16, 25, 27, 20, 9, 18, 13, 33, 35, 19, 18, 39, 41, 43, 28, 17, 47, 49, 51, 53, 55, 57, 59, 61, 39, 24, 65, 67, 69, 71, 35, 38, 75, 77, 79, 81, 47, 36, 85, 87, 89, 23, 68, 71, 10, 12, 95, 97, 99, 101, 103, 40, 65, 107, 109, 100
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| M. Waldschmidt writes: Conjecture 1.3 (Pillai). Let k be a positive integer. The equation x^p - y^q = k where the unknowns x, y, p and q take integer values, all =>2, has only finitely many solutions (x,y,p,q). This means that in the increasing sequence of perfect powers [A001597] the difference between two consecutive terms [the present sequence] tends to infinity. It is not even known whether for, say, k=2, Pillai's equation has only finitely many solutions. A related open question is whether the number 6 occurs as a difference between two perfect powers. See Sierpinksi [1970], problem 238a, p. 116. - Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 18 2008
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REFERENCES
| W. Sierpinski, 250 problems in elementary number theory, Modern Analytic and Computational Methods in Science and Mathematics, No. 26, American Elsevier, Warsaw, 1970, pp. 21, 115-116.
S. S. Pillai, On the equation 2^x - 3^y = 2^X - 3^Y, Bull, Calcutta Math. Soc. 37 (1945) 15-20.
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LINKS
| Daniel Forgues and T. D. Noe, Table of n, a(n) for n=1..10000
M. Waldschmidt, Open Diophantine problems
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FORMULA
| a(n) = A001597(n+1) - A001597(n). - Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 18 2008
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EXAMPLE
| Consecutive perfect powers are A001597[14]=121, A001597[13]=100, so a(13)=121-100=21
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CROSSREFS
| Cf. A053707, first differences of consecutive perfect prime powers.
Cf. A001597, A025475, A053707, A069623.
Sequence in context: A016607 A076446 * A076412 A053707 A075052 A111516
Adjacent sequences: A053286 A053287 A053288 * A053290 A053291 A053292
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KEYWORD
| nonn
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AUTHOR
| Labos E. (labos(AT)ana.sote.hu), Mar 03 2000
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